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satela [25.4K]
1 year ago
8

The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT. (a) At what distance is it 0.100μ

Physics
1 answer:
maw [93]1 year ago
4 0

The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

  • The magnitude and direction of the magnetic field due to a straight  wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by  B=\frac{u_0I}{2\pi r}  ...(1)
  • Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.

It is given that magnetic field 40.0 cm away from a straight wire is  1.00μT having  current 2.00 A .

From equation (1)  magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by    1.00=\frac{u_0I}{2\pi \times0.4}      .....(2)

From equation (1)  magnetic field 'r' m away from a straight wire is 0.100μT which is given by    0.100=\frac{u_0I}{2\pi \times r}       ...(3)

On dividing equation (2) by (3) , we get

             \frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m

Learn more about magnetic field here :

brainly.com/question/27939568

#SPJ4

 

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Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

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mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

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           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

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3 years ago
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The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

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I = I_{D} - I_{H} (1)

Where:

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Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

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m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

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2 years ago
A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
shusha [124]

Answer:

v=d\sqrt{\frac{k}{m}}

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U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

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the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

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matrenka [14]
<span> (a) does an electric field exert a force on a stationary charged object? 
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</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
6 0
3 years ago
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